# How do you find the limit of (x+1-cos(x))/(4x) as x approaches 0?

May 30, 2016

${\lim}_{x \to 0} \frac{x + 1 - C o s \left(x\right)}{4 x} = \frac{1}{4}$

#### Explanation:

We will be using

$C o s \left(a + b\right) = C o s \left(a\right) C o s \left(b\right) - S \in \left(a\right) S \in \left(b\right)$

and

${\sin}^{2} \left(a\right) + {\cos}^{2} \left(a\right) = 1$

Using $\cos \left(x\right) = 1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)$

and substituting

$\frac{x + 1 - C o s \left(x\right)}{4 x} = \frac{x + 2 {\sin}^{2} \left(\frac{x}{2}\right)}{4 x}$
${\lim}_{x \to 0} \frac{x + 2 {\sin}^{2} \left(\frac{x}{2}\right)}{4 x} = {\lim}_{x \to 0} \left(\frac{x}{4 x}\right) + {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{4 x}$

but

${\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{4 x} = {\lim}_{x \to 0} \sin \frac{\frac{x}{2}}{4} {\lim}_{x \to 0} \left(\sin \frac{\frac{x}{2}}{\frac{x}{2}}\right) = 0 \times 1$

so

${\lim}_{x \to 0} \frac{x + 1 - C o s \left(x\right)}{4 x} = \frac{1}{4}$